blocks|key|56400|text|使用分组。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|56401|foo.*?(foo)|code-block|syntax|javascript|56402|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Use grouping.

<pre><code>foo.*?(foo)
</code></pre>

blocks|key|55659|text|假设你想要的模式是abc%2Bd，你想要匹配这个模式在字符串中的第二次出现。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|55660|您将构造以下正则表达式：|55661|abc%2Bd.*?(abc%2Bd)|code-block|syntax|javascript|55662|这将匹配以下形式的字符串：<your-pattern>...<your-pattern>。既然我们使用了不情愿的限定符*？我们可以肯定的是，两者之间不会再有一场比赛了。使用几乎所有regex实现都提供的匹配器组，您可以检索带括号的组中的字符串，这正是您想要的。|offset|length|style|CODE|55663|entityMap^0|0|0|0|D|V|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|R|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|S|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|T|8|@$K|U|L|V|M|N]]|9|@]|A|$]]|$1|O|3|-4|5|6|7|W|8|@]|9|@]|A|$]]]|P|$]]

Suppose the pattern you want is abc+d. You want to match the second occurrence of this pattern in a string.

You would construct the following regex:

<pre><code>abc+d.*?(abc+d)
</code></pre>

This would match strings of the form: <code>&lt;your-pattern&gt;...&lt;your-pattern&gt;</code>. Since we're using the reluctant qualifier *? we're safe that there cannot be another match of between the two. Using matcher groups which pretty much all regex implementations provide you would then retrieve the string in the bracketed group which is what you want.

blocks|key|2543567|text|会不会像这样的东西|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2543568|(pattern.*?(pattern))*|code-block|syntax|javascript|2543569|为你工作？|2543570|编辑：|2543571|这样做的问题是它使用了非贪婪运算符*?，这可能需要沿字符串进行大量的回溯，而不是只查看每个字母一次。这意味着对于较大的差距，这可能会很慢。|offset|length|style|CODE|2543572|entityMap^0|0|0|0|0|H|2|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|U|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|V|8|@]|9|@]|A|$]]|$1|K|3|L|5|6|7|W|8|@$M|X|N|Y|O|P]]|9|@]|A|$]]|$1|Q|3|-4|5|6|7|Z|8|@]|9|@]|A|$]]]|R|$]]

Would something like 

<pre><code>(pattern.*?(pattern))*
</code></pre>

work for you?

Edit:

The problem with this is that it uses the non-greedy operator <code>*?</code>, which can require an awful lot of backtracking along the string instead of just looking at each letter once. What this means for you is that this could be slow for large gaps.

blocks|key|56440|text|如果你使用的是C#，你可以一次获取所有的匹配项(例如，使用Regex.Matches()，它返回一个MatchCollection，并检查项目的索引：index+%25+2+!=+0)。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|56441|如果您想要找到匹配项来替换它，请使用使用MatchEvaluator的Regex.Replace()重载之一(例如，Regex.Replace(String,+String,+MatchEvaluator)。代码如下：|56442|using+System;
using+System.Collections.Generic;
using+System.Text;
using+System.Text.RegularExpressions;

namespace+ConsoleApplication1
{
++++class+Program
++++{
++++++++static+void+Main(string[]+args)
++++++++{
++++++++++++string+input+=+"abcdabcd";

++++++++++++//+Replace+*second*+a+with+m

++++++++++++string+replacedString+=+Regex.Replace(
++++++++++++++++input,
++++++++++++++++"a",
++++++++++++++++new+SecondOccuranceFinder("m").MatchEvaluator);

++++++++++++Console.WriteLine(replacedString);
++++++++++++Console.Read();

++++++++}

++++++++class+SecondOccuranceFinder
++++++++{
++++++++++++public+SecondOccuranceFinder(string+replaceWith)
++++++++++++{
++++++++++++++++_replaceWith+=+replaceWith;
++++++++++++++++_matchEvaluator+=+new+MatchEvaluator(IsSecondOccurance);
++++++++++++}

++++++++++++private+string+_replaceWith;

++++++++++++private+MatchEvaluator+_matchEvaluator;
++++++++++++public+MatchEvaluator+MatchEvaluator
++++++++++++{
++++++++++++++++get
++++++++++++++++{
++++++++++++++++++++return+_matchEvaluator;
++++++++++++++++}
++++++++++++}

++++++++++++private+int+_matchIndex;
++++++++++++public+string+IsSecondOccurance(Match+m)
++++++++++++{
++++++++++++++++_matchIndex%2B%2B;
++++++++++++++++if+(_matchIndex+%25+2+==+0)
++++++++++++++++++++return+_replaceWith;
++++++++++++++++else
++++++++++++++++++++return+m.Value;
++++++++++++}
++++++++}
++++}
}|code-block|syntax|javascript|56443|entityMap^0|T|F|1E|F|23|E|0|K|E|Z|F|1M|19|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@$9|P|A|Q|B|C]|$9|R|A|S|B|C]|$9|T|A|U|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|V|8|@$9|W|A|X|B|C]|$9|Y|A|Z|B|C]|$9|10|A|11|B|C]]|D|@]|E|$]]|$1|H|3|I|5|J|7|12|8|@]|D|@]|E|$K|L]]|$1|M|3|-4|5|6|7|13|8|@]|D|@]|E|$]]]|N|$]]

If you're using C#, you can either get all the matches at once (i.e. use <code>Regex.Matches()</code>, which returns a <code>MatchCollection</code>, and check the index of the item: <code>index % 2 != 0</code>).

If you want to find the occurrence to replace it, use one of the overloads of <code>Regex.Replace()</code> that uses a <code>MatchEvaluator</code> (e.g. <code>Regex.Replace(String, String, MatchEvaluator)</code>. Here's the code:

<pre><code>using System;
using System.Collections.Generic;
using System.Text;
using System.Text.RegularExpressions;

namespace ConsoleApplication1
{
 class Program
 {
 static void Main(string[] args)
 {
 string input = "abcdabcd";

 // Replace *second* a with m

 string replacedString = Regex.Replace(
 input,
 "a",
 new SecondOccuranceFinder("m").MatchEvaluator);

 Console.WriteLine(replacedString);
 Console.Read();

 }

 class SecondOccuranceFinder
 {
 public SecondOccuranceFinder(string replaceWith)
 {
 _replaceWith = replaceWith;
 _matchEvaluator = new MatchEvaluator(IsSecondOccurance);
 }

 private string _replaceWith;

 private MatchEvaluator _matchEvaluator;
 public MatchEvaluator MatchEvaluator
 {
 get
 {
 return _matchEvaluator;
 }
 }

 private int _matchIndex;
 public string IsSecondOccurance(Match m)
 {
 _matchIndex++;
 if (_matchIndex % 2 == 0)
 return _replaceWith;
 else
 return m.Value;
 }
 }
 }
}
</code></pre>

blocks|key|55888|text|反向参考可以在这里找到有趣的解决方案。此正则表达式：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|55889|([a-z]%2B).*(\1)|code-block|syntax|javascript|55890|会找到最长的重复序列。|55891|这将找到一个重复的3个字母的序列：|55892|([a-z]{3}).*(\1)|55893|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@]|9|@]|A|$]]|$1|I|3|J|5|6|7|R|8|@]|9|@]|A|$]]|$1|K|3|L|5|D|7|S|8|@]|9|@]|A|$E|F]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

Back references can find interesting solutions here. This regex:

<pre><code>([a-z]+).*(\1)
</code></pre>

will find the longest repeated sequence.

This one will find a sequence of 3 letters that is repeated:

<pre><code>([a-z]{3}).*(\1)
</code></pre>

blocks|key|55717|text|没有“直接”的方法可以做到这一点，但是可以在：a[%5Ea]*a中指定两次与第二个"a“匹配的模式。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|55718|另一种选择是使用您的编程语言(perl？C#？...)来匹配第一次出现，然后第二次出现。|55719|编辑：我见过其他人使用“非贪婪”操作符进行响应，这可能是一个很好的方法，假设你的正则表达式库中有它们！|BOLD|55720|entityMap^0|N|7|0|0|0|2|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|P|8|@]|D|@]|E|$]]|$1|H|3|I|5|6|7|Q|8|@$9|R|A|S|B|J]]|D|@]|E|$]]|$1|K|3|-4|5|6|7|T|8|@]|D|@]|E|$]]]|L|$]]

There's no "direct" way of doing so but you can specify the pattern twice as in: <code>a[^a]*a</code> that match up to the second "a".

The alternative is to use your programming language (perl? C#? ...) to match the first occurence and then the second one.

EDIT: I've seen other responded using the "non-greedy" operators which might be a good way to go, assuming you have them in your regex library!

Is there a way to specify a regular expression to match every 2nd occurrence of a pattern in a string?

Examples

<ul>
<li>searching for a against string abcdabcd should find one occurrence at position 5</li>
<li>searching for ab against string abcdabcd should find one occurrence at position 5</li>
<li>searching for dab against string abcdabcd should find no occurrences</li>
<li>searching for a against string aaaa should find two occurrences at positions 2 and 4</li>
</ul>

Match at every second occurrence

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

文章

问答

视频

学习中心

腾讯云实验室

直播

竞赛

腾讯云代码分析专区

腾讯iOA零信任安全管理系统专区

腾讯云架构师技术同盟交流圈

腾讯云数据库专区

腾讯云顾问专区

腾讯云原生专区

腾讯混元专区

腾讯云TCE专区

腾讯云Lighthouse专区

腾讯云HAI专区

腾讯云Edgeone专区

腾讯云存储专区

腾讯云智能专区

腾讯轻联专区 

腾讯云开发专区

TAPD专区

腾讯轻量云游戏服专区

腾讯云最具价值专家

腾讯云架构师技术同盟

腾讯云创作之星

腾讯云开发者先锋

腾讯云代码助手

云原生构建

TAPD 敏捷项目管理

Cloud Studio

SDK中心

API中心

命令行工具

涵盖代码开发、场景应用、自动测试全流程，助你从零构建专属AI助手

一站式MCP教程库，解锁AI应用新玩法

有没有一种方法可以指定一个正则表达式来匹配字符串中每隔两次出现的模式？示例搜索a 字符串abcdabcd应在位置5处找到一个匹配项搜索ab E29应在位置5处找到一个匹配项abcdabcd E211应在位置5处找到一个匹配项H212</code><代码>H113应在位置5处找到一个匹配项搜索e114</code>dab e215</code>应在位置2和4处找到两个匹配项E116aaaaa应在位置

问每隔一秒匹配一次
EN

回答 6

Stack Overflow用户

Stack Overflow用户

Stack Overflow用户

社区

活动

圈层

关于

腾讯云开发者

热门产品

热门推荐

更多推荐

问每隔一秒匹配一次EN

回答 6

Stack Overflow用户

Stack Overflow用户

Stack Overflow用户

社区

活动

圈层

关于

腾讯云开发者

热门产品

热门推荐

更多推荐

问每隔一秒匹配一次
EN